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  • 【HDOJ】4373 Mysterious For

    1. 题目描述
    有两种不同类型的循环,并给出一个由1、2组成的序列,表示嵌套的循环类型。
    问这样组着的循环一共需要多少次循环?并将结果模364875103。

    2.基本思路
    显然,每当遇到一个类型1的序列,即可以判定12...2的嵌套循环共多少次,而1类型的循环次数为常亮。
    因此,将原序列从1分开,并将每个子序列的循环次数相乘即为总的循环次数。
    1     共循环n次 = C[n][1]
    12   共循环n*(n+1)/2次 = C[n+1][2]
    122 共循环n*(n+1)*(n+2)/6次 = C[n+2][3]
    12..2 |2|=m 共循环n*(n+1)*(n+2)/6次 = C[n+m-1][m]
    由Lucas定理可知,假设p为素数,有

    比较悲剧的是364875103是个合数,将它拆解为两个素数并使用LUCAS后,
    我们可以知道ans=a1(mod p1), ans=a2(mod p2), 且p1, p2互素,求ans = ? (mod p1*p2)。
    使用中国剩余定理。
    使用欧拉定理求逆。

    3. 代码

      1 /* 4373 */
      2 #include <iostream>
      3 #include <sstream>
      4 #include <string>
      5 #include <map>
      6 #include <queue>
      7 #include <set>
      8 #include <stack>
      9 #include <vector>
     10 #include <deque>
     11 #include <algorithm>
     12 #include <cstdio>
     13 #include <cmath>
     14 #include <ctime>
     15 #include <cstring>
     16 #include <climits>
     17 #include <cctype>
     18 #include <cassert>
     19 #include <functional>
     20 #include <iterator>
     21 #include <iomanip>
     22 using namespace std;
     23 //#pragma comment(linker,"/STACK:102400000,1024000")
     24 
     25 #define sti                set<int>
     26 #define stpii            set<pair<int, int> >
     27 #define mpii            map<int,int>
     28 #define vi                vector<int>
     29 #define pii                pair<int,int>
     30 #define vpii            vector<pair<int,int> >
     31 #define rep(i, a, n)     for (int i=a;i<n;++i)
     32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     33 #define clr                clear
     34 #define pb                 push_back
     35 #define mp                 make_pair
     36 #define fir                first
     37 #define sec                second
     38 #define all(x)             (x).begin(),(x).end()
     39 #define SZ(x)             ((int)(x).size())
     40 #define lson            l, mid, rt<<1
     41 #define rson            mid+1, r, rt<<1|1
     42 #define LL                __int64
     43 
     44 const int maxn = 25;
     45 const int mod1 = 97;
     46 const int mod2 = 3761599;
     47 const int mod = 364875103;
     48 LL fact1[mod1+15];
     49 LL fact2[mod2+15];
     50 LL e1, e2;
     51 int pos[maxn];
     52 int n, m, k;
     53 
     54 LL Pow(LL base, LL n, LL mod) {
     55     LL ret = 1;
     56     
     57     base %= mod;
     58     while (n) {
     59         if (n & 1)
     60             ret = ret * base % mod;
     61         base = base * base % mod;
     62         n >>= 1;
     63     }
     64     
     65     return ret;
     66 }
     67 
     68 void init() {
     69     fact1[0] = fact2[0] = 1;
     70     rep(i, 1, mod1)
     71         fact1[i] = fact1[i-1] * i % mod1;
     72     rep(i, 1, mod2)
     73         fact2[i] = fact2[i-1] * i % mod2;
     74     e1 = mod2 * Pow(mod2, mod1-2, mod1);
     75     e2 = mod1 * Pow(mod1, mod2-2, mod2);
     76     
     77     #ifndef ONLINE_JUDGE
     78         printf("e1 = %I64d, e2 = %I64d
    ", e1, e2);
     79     #endif
     80 }
     81 
     82 LL C(LL n, LL m, LL mod, LL *fact) {
     83     if (n < m)    return 0;
     84     return fact[n] * Pow(fact[m]*fact[n-m], mod-2, mod) % mod;
     85 }
     86 
     87 LL Lucas(LL n, LL m, LL mod, LL *fact) {
     88     if (m == 0)    return 1;
     89     return C(n%mod, m%mod, mod, fact) * Lucas(n/mod, m/mod, mod, fact);
     90 }
     91 
     92 void solve() {
     93     LL ans = 1, tmp;
     94     LL a1, a2;
     95     
     96     rep(i, 0, k) {
     97         m = pos[i+1]-pos[i];
     98         a1 = Lucas(n+m-1, m, mod1, fact1);
     99         a2 = Lucas(n+m-1, m, mod2, fact2);
    100         tmp = (a1*e1 + a2*e2) % mod;
    101         #ifndef ONLINE_JUDGE
    102             printf("a1 = %I64d, a2=%I64d, tmp=%I64d
    ", a1, a2, tmp);
    103         #endif
    104         ans = ans * tmp % mod;
    105     }
    106     
    107     printf("%I64d
    ", ans);
    108 }
    109 
    110 int main() {
    111     ios::sync_with_stdio(false);
    112     #ifndef ONLINE_JUDGE
    113         freopen("data.in", "r", stdin);
    114         freopen("data.out", "w", stdout);
    115     #endif
    116     
    117     int t;
    118     
    119     init();
    120     scanf("%d", &t);
    121     rep(tt, 1, t+1) {
    122         scanf("%d%d%d", &n, &m, &k);
    123         rep(i, 0, k)
    124             scanf("%d", &pos[i]);
    125         pos[k] = m;
    126         printf("Case #%d: ", tt);
    127         solve();
    128     }
    129     
    130     #ifndef ONLINE_JUDGE
    131         printf("time = %d.
    ", (int)clock());
    132     #endif
    133     
    134     return 0;
    135 }


    4. 数据生成器

     1 from copy import deepcopy
     2 from random import randint, shuffle
     3 import shutil
     4 import string
     5 
     6 
     7 def GenDataIn():
     8     with open("data.in", "w") as fout:
     9         t = 20
    10         bound = 10**5
    11         fout.write("%d
    " % (t))
    12         for tt in xrange(t):
    13             n = randint(20, bound)
    14             m = randint(20, bound)
    15             k = randint(1, 15)
    16             fout.write("%d %d
    %d " % (n, m, k))
    17             L = [0]
    18             st = set()
    19             for i in xrange(1, k):
    20                 while True:
    21                     x = randint(1, m)
    22                     if x not in st:
    23                         break
    24                 L.append(x)
    25                 st.add(x)
    26             L = sorted(L)
    27             fout.write(" ".join(map(str, L)) + "
    ")
    28             
    29             
    30 def MovDataIn():
    31     desFileName = "F:eclipse_prjworkspacehdojdata.in"
    32     shutil.copyfile("data.in", desFileName)
    33 
    34     
    35 if __name__ == "__main__":
    36     GenDataIn()
    37     MovDataIn()
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/5204942.html
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